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mixer [17]
3 years ago
15

Why were the French and Dutch colonized areas so small compared to the Spanish colonized areas?

Engineering
2 answers:
Degger [83]3 years ago
4 0
The French and Dutch colonized arenas so small compared to the Spanish colonized areas For farming
lara31 [8.8K]3 years ago
4 0
Because of farming tahts why they are so small
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Solid Isomorphous alloys strength
Sphinxa [80]

Answer:

Explanation:

ℎ

3 0
3 years ago
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
taurus [48]

Answer:

<em>No, the velocity profile does not change in the flow direction.</em>

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

3 0
2 years ago
If block A of the pulley system is moving downward at 6 ft&gt;s while block C is moving down at 18 ft&gt;s, determine the relati
kramer

Answer:

Explanation:

The detailed steps and appropriate calculation with analysis is as shown in the attachment.

3 0
2 years ago
CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic
valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd2)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd3)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd4)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

p = fork();

if (p < 0)

{

 fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

 close(fd1[0]);// Close reading end of first pipe

 char concat_str[100];

 printf("\n\tEnter meaaage:"):

 scanf("%s",concat_str);

 write(fd1[1], concat_str, strlen(concat_str)+1);

 // Concatenate a fixed string with it

 int k = strlen(concat_str);

 int i;

 for (i=0; i<strlen(fixed_str); i++)

 {

  concat_str[k++] = fixed_str[i];

 }

 concat_str[k] = '\0';//string ends with '\0'

 // Close both writting ends

 close(fd1[1]);

 wait(NULL);

//.......................................................................

 close(fd2[1]);

 read(fd2[0], concat_str, 100);

 if(strcmp(concat_str,"invalid")==0)

 {

 printf("\n\tmessage not send");

 }

 else

 {

  printf("\n\tmessage send to prog_2(child_2).");

 }

 close(fd2[0]);//close reading end of pipe 2

 exit(0);

}

else

{

 close(fd1[1]);//Close writting end of first pipe

 char concat_str[100];

 read(fd1[0], concal_str, strlen(concat_str)+1);

 close(fd1[0]);

 close(fd2[0]);//Close writing end of second pipe

 if(/*check if msg is valid or not*/)

 {

  //if not then

  write(fd2[1], "invalid",sizeof(concat_str));

  return 0;

 }

 else

 {

  //if yes then

  write(fd2[1], "valid",sizeof(concat_str));

  close(fd2[1]);

  q=fork();//create chile process 2

  if(q>0)

  {

   close(fd3[0]);/*close read head offd3[] */

   write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

   close(fd3[1]);

   wait(NULL);//wait till child_process_2 send ACK

   //...........................................................

   close(fd4[1]);

   read(fd4[0],concat_str,100);

   close(fd4[0]);

   if(sctcmp(concat_str,"ack")==0)

   {

    printf("Messageof child process_1 is received by child process_2");

   }

   else

   {

    printf("Messageof child process_1 is not received by child process_2");

   }

  }

  else

  {

   if(p<0)

   {

    printf("Chiile_Procrss_2 not cheated");

   }

   else

   {

     

    close(fd3[1]);//Close writing end of first pipe

    char concat_str[100];

    read(fd3[0], concal_str, strlen(concat_str)+1);

    close(fd3[0]);

    close(fd4[0]);//Close writing end of second pipe

    write(fd4[1], "ack",sizeof(concat_str));

     

   }

  }

 }

 close(fd2[1]);

}

}

8 0
3 years ago
A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug
Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

\rho = \frac{m}{V}

Where

m = Mass

V = Volume

For state one we know that

\rho_1 = \frac{m_1}{V}

m_1 = \rho_1 V

m_1 = 1.18*1

m_1 = 1.18Kg

For state two we have to

\rho_2 = \frac{m_2}{V}

m_2 = \rho_2 V

m_1 = 7.2*1

m_1 = 7.2Kg

Therefore the total change of mass would be

\Delta m = m_2-m_1

\Delta m = 7.2-1.18

\Delta m = 6.02Kg

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0
2 years ago
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